Question

A sample of semiconductor has a cross sectional area of 1cm^2 and a thickness of 0.1cm.

(a) Determine the number of electron hole pairs that are generated per unit volume per unit time by the uniform absorpotion of 1 Watt of light at a wavelength of 6300 A. Assume each photon creates one electron hole pair.
(b) If the excess minority carrier lifetime is 10μs, what is the steady state excess carrier concentration?

Answer 1

a. 3.17*10¹⁹ b. 3.17*10¹⁴

Step-by-step explanation:

a. Area A = 1cm²

Thickness h = 0.1cm

Energy of photon E = hf=hc/λ

Where f = frequency; c=speed of light; λ=wavelength = 6300Α=6300*10⁻¹⁰;

Planck's constant h = 6.626*10⁻³⁴ joule-seconds; speed of light c = 3*10⁸

Therefore E = (6.626*10⁻³⁴)*3*10⁸/6300*10⁻¹⁰=3.155*10¹⁹J

1Watt of light releases 3.17*10¹⁸ photons per second.

Volume of sample = Area * Thickness = 1*0.1=0.1cm³

Therefore, number of electron hole pairs that are generated per unit volume per unit time = 3.17*10¹⁸/0.1 = 3.17*10¹⁹photon/cm³-s

b. Steady state excess carrier concentration = 3.17*10¹⁹*10μs

=3.17*10¹⁹*10*10⁻⁶

=3.17*10¹⁴/cm³